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$$f(x) = \left\{\begin{array}{lcl}-x^3 + 1&\text{if}&x\geq0\\ -x^2+2x&\text{if}&x<0\end{array}\right.$$ I want to find the critical points but after differentializing and equalizing to $0$, the $x$ which one is 1 and other is 0. (I think that the point should be the same one?)

What's wrong with it?

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The function isn't even continuous at $\,x=0\,$ , so it cannot be derivable there. Thus, as the Babak points out, $\,x=0\,$ is automatically a critical point per definition. –  DonAntonio Dec 25 '12 at 13:36
    
@DonAntonio That's a strange definition, at least in Italy: a critical point, for us, is a point where the derivative exists and is zero. So you usually teach that $0$ is a critical point of $|\cdot|$? –  Siminore Dec 25 '12 at 14:45
    
@Siminore, for a moment you made me doubt, but it is in fact a rather widespread definition. you can check it in en.wikipedia.org/wiki/Critical_point_(mathematics) , or in the book by Thomas (chapter 4.1), or here tinyurl.com/cg6nat3 , or here tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx , or chapter 3.1 of the Thomas-Finney book...And yes: $\,0\,$ is a critical point of the absolute value function. –  DonAntonio Dec 25 '12 at 15:02
    
@DonAntonio Yes, it seems quite widespread in calculus textbooks, at least in the US. But Spivak (Calculus, chapter 11) says that a point is critical if $f'(x)=0$. Here we tend to classify points of non-differentiability, but wwe reserve the name critical or stationary to the zeroes of the first derivative. –  Siminore Dec 25 '12 at 17:08

2 Answers 2

up vote 3 down vote accepted

Hint: We can see that $x=0$ is a critical point for the function because $$f'^+(0)=-3(0^2)\neq 2=-2(0)+2=f'^-(0) $$

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If I'm not mistaken, this shows that $f$ is not continuously differentiable at $x = 0$, but it could still be differentiable there. –  Michael Albanese Dec 25 '12 at 13:43
    
$\quad \ddot\smile\quad$ –  amWhy Mar 3 '13 at 0:13

If I understand the post correctly, you view $x=1$ as a critical point of $f(x)$. It isn't. Although certainly the derivative of $-x^2+2x$ is $0$ at $x=1$, the function $f(x)$ is $-x^2+2x$ for negative $x$ only.

As to whether $0$ should be called a critical point, that depends on the definition being used in your course. The usual convention in elementary North American texts would include points of discontinuity of the function, and points where the derivative does not exist.

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